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Help solving third order pde?

  1. Oct 12, 2009 #1
    2Uxxy+3Uxyy-Uxy=0 where U=U(x,y)

    I made the substitution W=Uxy and then used a change of coordinates (n= 2x+3y, and r=3x-2y) which reduced the problem to solving Uxy=f(3x-2y)exp((2x+3y)/3) because W=f(r)exp(n/3). Now I have no idea where to go from there. Any help would be much appreciated.

    Thanks
     
  2. jcsd
  3. Oct 13, 2009 #2
    The general solution to your PDE is as follows

    U(x,y) = F1(x)+F2(y)+exp(x/2)F3(3x-2y),

    where F1, F2, F3 are arbitrary functions.
     
  4. Oct 13, 2009 #3
    kosovtsov,

    Would you mind explaining to me how you arrived at that solution? It would be greatly appreciated.

    Thanks
     
  5. Oct 14, 2009 #4
    There is a general principle. If a problem can be solved, it as a rule, can be solved by countless number of methods.

    For example, from your

    Uxy=f(3x-2y)exp((2x+3y)/3)

    it is follows immediately by double integration that the general solution can be in form

    U(x,y)=F1(x)+F2(y)+\int \int (f(3x-2y)exp((2x+3y)/3)) dx dy .

    My solution only looks simpler, but is equivalent the one above.
     
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